# Probability of getting at least 1 white ball

There's 1 box with 10 balls:

3 white 2 black 5 blue

2 balls are removed the box randomly, and the first one is not replaced.

What is the probability of getting at least 1 white ball?

Here's what I cooked up:

The probability of getting at least 1 white ball equals the sum of the probabilities of getting a white ball first and getting a white ball second. The probability of getting a white ball on the first one is 3/10. The probability of getting a white ball on the second one is 2/9.

So the probability would be:

3/10 + 2/9 = 47/90

I'm solving this on a book which says the answer is 8/15.

Am I doing anything wrong?

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What if the first ball is not white? – Lord_Farin Oct 20 '12 at 11:38
Then it'd be 3/10 + 3/9, and that is also not 8/15... – David Gomes Oct 20 '12 at 11:42
The probability of the second ball being white is not independent of whether or not the first ball is white. Thus, your summation clause does not apply. (And even if it did, you would still have to correct for the case that both occurred, which you then counted twice.) – Lord_Farin Oct 20 '12 at 11:43
If you apply your argument to the case of drawing three balls, each of which is blue, you'd get $95/72$.. – DSM Oct 20 '12 at 11:49

Here Question is What is the probability of getting at least 1 white ball ? So there are 3 possible cases as follows

1. 1st is white AND 2nd is not for that (3/10) * (7/9)

OR 2. 1st is not white AND 2nd is white for that (7/10)*(3/9) (adding dummy characters in order to save the edit)

OR 3. both are white for that (3/10) * (2/9)

In probability questions remember OR means ADD, AND means Multiply

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What's that thing about punctuation again? Ah, I recall. It makes a piece of text readable... – Lord_Farin Oct 20 '12 at 11:41
now answer is clear to you? – madwhoo Oct 20 '12 at 11:46
Yes, but it still misses punctuation. Take your time to properly write your answer; it is more satisfactory and useful if you do. – Lord_Farin Oct 20 '12 at 11:48
this is my 1st day on this website. i will improve – madwhoo Oct 20 '12 at 11:53
Which is precisely why I have given you the advice :). Have a good time. PS: The advice also applies to comments; I'm sure you'll get it soon enough. – Lord_Farin Oct 20 '12 at 11:57

You want to know the probability of "one of the two balls is white" = "the first ball is white OR (the first ball isn't white AND the second ball is white)".

OR is +

AND is $\times$

So you get $\frac{3}{10} + \frac{10-3}{10}\times \frac{3}{10-1}$.

$\frac{3}{10}$ and $\frac{10-3}{10}$ because you have 3 white balls and 10 balls so 10 - 3 not-white balls

$\frac{3}{10-1}$ because when you pick the second ball, you removed one non-white ball

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or you can arrive at the same answer by thinking in reverse. If at least one ball is white, then the only other combination is Non-white in the 1st pick and No-white in the 2nd pick. Then 1-(2 non-white probability)=at least 1 white picked!

So 1st pick for non-white = 7/10 and the 2nd pick for non-white = 6/9. This probability for this combination is 7/10 x 6/9 = 7/15. So the probability of getting at least 1 white = 1-7/15= 8/15 = 0.53333....

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